Pratap technical experience

  Question= A power station having a maximum demand of 100 mw has a load factor of 30% and is to be supplied by one of the following schemes. (1) A steam station in conjunction with a hydro - electric station, the latter supplying 100×10⁶ kwh per annum with a maximum output of 40 mw.  (2) a steam station capable of supplying the whole load .  (3) a hydro station capable of supplying the whole load. Compare the overall cost per kwh generated ,??  assuming the following data . Solution= overall cost/kwh= (total annual charges for both steam and hydro station)÷ ( unit generated per annum) ,  » Total annual charges steam station =   • so, steam station , total annual cost= operating cost+annual interest and depreciation,  » operating cost= operating cost/kwh 5 paise × unit supplied by station,  » unit supplied by station = unit generated per annum - unit supplied by hydro station ,  » unit generated per annum = maximum demand × L F. × hours...

  Question= Compare the annual cost of supplying a factory load having a maximum demand of 1 megawatt and a load factor of 50% by energy obtained from (1) a private oil engine generating plant ? and (2) public supply ?  (1) detail of private oil engine generating unit =capital cost=RS 12×10⁵ , cost of repair and maintenance = RS 0.005 per kwh generated , cost of fule = RS 1600 per 1000 kg , interest and depreciation = 10% per annum, fule consumption = 0.3 kg/kwh generated, wages = RS 50,000 per annum,, (365days×24 hours = 8760 hours)  (2) public supply company , RS 150 per kw of maximum demand plus 15 paise per kwh.  Solution=  1 st compare = total annual running charges = annual cost of fuel + annual cost of repair and maintenance + annual wages +  annual interest and depreciation . • annual cost of fuel = annual fule consumption × RS 1600 per 1000 kg ,  » annual fule consumption = unit generated per annum  × fuel consumption (0.3 kg/kwh) ,...

  Question = A hydro electric plant costs RS 3000 per kW of installed capacity . The total annual charges consist of 5 % (F) as interest , depreciation at 2% (F) , operation and maintenance at 2 % (R) and insurance , rent etc , 1.5 % (R). Determine a suitable two part tariff if the losses in transmission and distribution are 12.5% and diversity of loads is 1.25 . Assume that maximum demand on the station is 80% of the capacity and annual load factor is 40% . What is the overall cost of generation per kWh ? Solution = overall cost /kWh = total annual charges ÷ unit reaching the consumer ,  » Total annual charges = annual fixed charges + annual running charges ,  » annual fixed charges (A F.C ) = capital cost of plant × ( interest % + depreciation % ) /100 , » capital cost of plant = let installed capacity of the station kW × plant cost per kW , » 100 kW × 3000 RS = 3 × 10⁵ RS ,  » interest % + depreciation % ) /100 = (5 +2 ) ÷ 100  » So , A F.C. = 3 × 10⁵ RS ...

  Question = The annual working Cost of a power station is represented by the formula RS ( a+b kW+c kwh ) where the various terms have their usual meaning . Determine the value of a , b and c for a 60 MW station operating at annual load factor of 50% from the following data :  • F = fixed charges , R = running charges ,  (1) capital cost of building and equipment is RS 5 × 10⁶ ( 50 lakh ) ,  (2) the annual cost of fuel , oil , taxation and wages of operating staff is RS 9,00,000 ( 9 lakhs ) . (R) (3) the interest and depreciation on building and equipment are 10% per annum , (F) (4) annual cost of organisation and interest on cost of site etc is RS 5,00,000 ( 5 lakhs ) . (F)  Solution = is given ,  » annual operating cost = RS ( a + b kw + c kwh ) ,   » a = annual fixed cost , • b = b × kW = annual semi fixed cost , • c =  c × kWh =  annual running cost ,  (1) a = annual fixed cost = the annual fixed cost is due to the annual cost o...

  Question = At the end of a power distribution system , a certain feeder supplies three distribution transformers ,each one supplying a group of customer whose connected load are as under :  • transformer no. 1 load 10 kw , demand factor 0.65 , diversity of groups 1.5 , • transformer no. 2 load 12 kw , demand factor 0.6 , diversity of groups 3.5 , • transformer no. 3 load 15 kw , demand factor 0.7 , diversity of groups 1.5 , If the diversity factor among the transformer is 1.3 , find the maximum load on the feeder ?  Solution = Show a feeder supplying three distribution transformers .  » formula = maximum demand or , load  on feeder = sum of maximum demand transformer 1 + transf 2 +trans 3  ÷ diversity factor ,  » So , maximum demand trans.. no. 1 = ( connected load of trans. 1 × demand factor ) ÷ diversity factor  ,  = ( 10 kW × 0.65 ) ÷ 1.5 , = 4.33 kw ,  » maximum demand of trans . 2 =  ( 12 kw × 0.7 ) ÷ 3.5 , ...

  Question = The capital cost of a hydro - power station of 50 MW capacity , is RS 1000 per kW. The annual depreciation charges are 10% (F) of the capital cost. A royalty of re 1 per kW per year and re 0.01 per kWh generated is to be paid for using the river water for generation of power. The maximum demand on the power station is 40 MW and annual load factor is 60 . Annual cost of salaries, maintenance charges etc is RS 700000. If 20% of the expense is also chargeable as fixed charges , calculate the generation cost in two part form ? Solution = 1 st part, annual running charges = formula ,  » Cost per kW = cost per kW due to fixed charges + royalty ,  » given, royalty = 1 RS ,  » cost per kW due to fixed charges  = total annual fixed charges ÷ M.D.  » is M.D. = 40 × 10³ kW ,  » Total annual fixed charges = depreciation charges of capital cost + charges of salaries , maintenance etc ,  » capital cost of plant = 50 × 10³ × 1000 , = 50 × 10⁶ ...

  Question = A generating station has the following data : installed capacity = 300 MW , capacity factor = 50% or 0.5 , annual load factor = 60% or 0.6 , annual cost of fuel , oil etc (R) =  RS 9×10⁷ , capital cost (F) = RS 10⁹ , annual interest and depreciation (F) = 10% ,  • Calculate (1) the minimum reserve capacity of the station and (2) the cost per kWh generated ? Solution = reserve capacity = installed capacity - maximum demand ,  » is given , Installed capacity = 300 MW ,  » M.D. = ( capacity factor × installed capacity ) ÷ Load factor ,  » capacity factor ( c f. ) = Average demand (A.D.) ÷ Installed capacity (I C.) ,  equations (1) AND ,  Load factor = average demand ( A.D.) ÷ maximum demand ( M.D.) , equations (2) » Dividing 1 and 2 we get ,  » C.F. ÷ L.F. = ( A.D. ÷ I.C. ) ÷ ( A.D. ÷ M.D.) , » after calculate = C.F. ÷ L.F. = M.D. ÷ I.C. , » So , Maximum demand = ( C.F. × I.C. ) ÷ L.F. ,  » M D.= ( 0.5 × 300 MW ) ÷ 0.6 , =...

  Question =Estimate the generating cost per kWh delivered from a generating station from the following data :  » F = fixed charge , R = running charge ,  Plant capacity = 50 MW , annual load factor = 40% , capital cost (F) = 1.2 crores , annual cost of wages , taxation etc (F) = RS 4 lakhs , Cost of fuel , lubrication , maintenance etc (R) = 1 paise/kWh generated . Interest 5% per annum (F) per annum , depreciation 6% per annum of initial value .  Solution = is given ,  » maximum demand and plant capacity = 50 MW , or 50 × 10³ kWh , • capital cost = 1.2 crores , or 120 × 10⁵ ,  • annual cost of wages , taxation etc (F) = RS 4 lakhs or RS 4 × 10⁵  , • convert % to point = 5% = 0.05 , 6% = 0.06 , • L F. = 40 % = 0.4 , • 1 paise = 1÷ 100 = 0.01 RS ,  » cost per unit = total annual charges ÷ unit generated per annum ,  » Total annual charges = cost of fuel , lubrication etc (R) + total annual fixed charges ,  »  cost of fuel , lubri...

Question =  A generating Plant has a maximum capacity of 100 kW and cost RS 1,60,000 . The annual fixed charges are 12% consisting of 5% intererst , 5% depreciation and 2% taxes . Find the fixed charges per kWh if the load factor is (1) 100% (2) 50% .  Solution = (1) load factor is 100% ,  » fixed charges per kWh = annual fixed charge ÷ unit generated per annum ,  » annual fixed charges = 1,60,000 × RS 0.12 = 19,200 RS ,  » unit generated per annum = maximum demand × L.F. × hours in a years , = 100 kW × 1 × 8760 ,= 8,76,000  kWh ,  » So, fixed charges per kWh = RS 19,200 ÷ 8,76,000 = 0.0219 RS or 2.19 paise , answer ✓  (2) load factor is 50% ,  »fixed charges per kWh = annual fixed charge ÷ unit generated per annum ,  » annual fixed charges = 1,60,000 × RS 0.12 = 19,200 RS ,  » unit generated per annum = maximum demand × L.F. × hours in a years , = 100 kW × 0.5 × 8760 , = 4,38,000 kWh ,  » So, fixed charges per kWh = RS 19...

  Question = A generating station has an installed capacity of 50,000 kW and delivers 220 × 10⁶ units per annum . If the annual fixed charges are RS 160 per kW installed capacity and running charges are 4 paise per kWh , determine the Cost per unit generated ? Solution = Cost per unit generated = total annual charges ÷ unit per annum ,  » is given , unit per annum = 220 × 10⁶ kWh ,  » Total annual charges = annual fixed charges + annual running charge ,  » annual fixed charges = 160 × plant capacity , = 160 RS × 50,000 kW = 80 × 10⁵ RS ,  » annual running charge = ( 0.04 RS ) × ( 220  × 10⁶ ) kWh = 88 × 10⁵ RS ,  » So , Total annual charges = ( 80 × 10⁵ ) + ( 88 × 10⁵ ) = 168 × 10⁵ RS ,  » now , Cost per unit generated = ( 168 × 10⁵ RS ) ÷ ( 220 × 10⁶ ) = 0.0764 RS or 7.64 paise , answer ✓  

  Question = A generating station has a maximum demand of 50,000 KW . Calculate the cost per unit generated from the following data : capital cost = RS 95 × 10⁶ , annual load factor = 40 % , annual cost of fuel and oil (R) = RS 9 × 10⁶ , interest and depreciation (F) = 12 % , taxes , wages and salaries etc (R) = RS 7.5 ×10⁶ ,  Where , F = fix charge , R = running charge ,  Solution = Cost per unit generated = total annual fix charge ÷ unit generated per annum ,  » unit generated per annum = max..  demand × load factor × hours in a years , = 50,000  kw × 40 % × 8760 hours = 17.52 × 10⁷ kWh ,  » annual fixed charge = (1) annual interest and depreciation = 12% of capital cost = 95 × 10⁶ × 0.12 = 11.4 × 10⁶ RS ,  ( 2nd ) annual running charge = total annual running charge  = annual Cost of  oil and fule + taxes wages etc , = ( 9 × 10⁶ + 7.5 × 10⁶ ) = 16.5 ×10⁶ RS ,   » add charge = ( 11.4 × 10⁶ + 16.5 ×10⁶ ) = 27.9 ×10⁶ ,  ...

  Question = A transformer costing RS 90,000 has a useful life of  20 years . Determine the annual depreciation ( मूल्यहास ) charge  ( RS )using straight line mathod . ? Assume the salvage value of the equipment to be 10,000 . Solution = is given , » Initial cost of transformer , P = 90,000 RS ,  » useful life n = 20 years ,  » salvage value , s = 10,000 rs ,  » annual depreciation charge = ( p- s )÷ n ,  » So , ( 90,000 - 10,000 ) ÷ 20 , = 80,000 ÷ 20 , = 4000 , RS answer ✓

  Question = The annual load duration curve for typical heavy load being served by a steam station , a ran - of - river station and a reservoir hydro - electric station is as shown in figure . The ratio of number of unit supplied by these  station is as follows : Steam : run - of - river : reservoir : 7:4:1   The run - of - river station is capable of generating power continuously and work as a base load station . The reservoir station works as a peak load station . Determine (1) the maximum demand on each station ? And (2)  load factor of each station ?  » solution =  » ODCA is the annual load duration curve for the system as shown in figure . The energy supplied by the reservoir Plant (1)  is represented by area DFG , steam station (2) by area FGCBE and run - of - river ( part 3 ) by area OEBA . The maximum and minimum loads on the system are 320 MW and 160 MW respectively . » 24 hours × 365 days = 8760 hours ,  » Unit ge...